Chapter 2. Divisibility. 2.1 Common Divisors
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1 Chapter 2 Divisibility 2.1 Common Divisors Definition Let a and b be integers. A common divisor of a and b is any integer that divides both a and b. Suppose that a and b are not both zero. By Proposition 1.3.8, there exists a greatest common divisor, which is positive. It is denoted by gcd(a, b). Note that in the proof of Proposition we proved that, for integers a and b which are not both zero, the set S = {x Z : x a, x b} has a maximum. Thus, gcd(a, b) = max{x Z : x a, x b}. Example The common divisors of 12 and 30 are: 6, 3, 2, 1, 1, 2, 3, 6. These are also the common divisors of 12 and 30, and of 6 and 0. Hence, gcd(12, 30) = gcd( 12, 30) = gcd(6, 0) = 6. Theorem Let a and b be integers, not both of which are zero. Then there exist integers x and y such that ax + by = gcd(a, b). Example Verifying Theorem in one case: 12( 2) + 30(1) = gcd(12, 30). Proposition Let a and b be integers, not both zero. Then any common divisor of a and b is a divisor of gcd(a, b). Definition Let a and b be positive integers. By Proposition 1.3.9, there exists a least common positive multiple of a and b. It is denoted by lcm(a, b). It follows from the proof of Proposition that for positive integers a and b lcm(a, b) = min{x Z : x > 0, a x, b x}. 13
2 14 June 25, :53 Chapter 2 Divisibility Proposition Let a and b be positive integers. Then any common multiple of a and b is a multiple of lcm(a, b). Proposition Let a, b and d be positive integers. If d is a common divisor of a and b, then ab/d is a common multiple of a and b. Proposition Let a, b and m be positive integers. If m is a common multiple of a and b and m divides ab, then ab/m is a common divisor of a and b. Proposition If a and b are positive integers, then ab = gcd(a, b) lcm(a, b). 2.2 Relatively Prime Integers Definition Let a and b be integers, not both zero. If gcd(a, b) = 1, then a and b are said to be relatively prime. Notice that the only common divisors of relatively prime integers are 1 and 1. Proposition Let a and b be integers that are not relatively prime. Then there exists a prime that divides both a and b. Proposition Let a and b be relatively prime positive integers and let c be an integer. If a c and b c, then ab c. Lemma (Euclid s Lemma). Let a and b be relatively prime integers. If c is an integer and a bc, then a c. Proposition Let a and b be integers, not both zero. If d = gcd(a, b), then a/d and b/d are relatively prime. Proposition Let p be a prime and a an integer. If p does not divide a, then p and a are relatively prime. Theorem (Euclid s First Theorem). Let p be a prime and let a and b be integers. If p ab, then p a or p b. Proposition Let a 1, a 2,...,a n be integers and let a = a 1 a 2 a n. If p is a prime and p divides a, then there exists k such that p divides a k. Proposition Let a 1, a 2,...,a n be integers and let a = a 1 a 2 a n. If b is a nonzero integer and b is relatively prime to each a k, then a and b are relatively prime. Proposition Let a 1, a 2,...,a n be integers greater than 1 and let a = a 1 a 2 a n. Then a + 1 is not divisible by any of the a k. Theorem (Euclid s Second Theorem). There exist infinitely many primes.
3 Section 2.3 Factoring Integers into Primes June 25, : Factoring Integers into Primes Proposition Let a be an integer greater than 1. Then there exists a positive integer n and there exist primes p 1, p 2,...,p n such that p 1 p 2 p n and a = p 1 p 2 p n. Theorem (The Fundamental Theorem of Arithmetic). Any integer greater than 1 has a unique representation as a product of primes in their natural order. Definition Let a be an integer greater than 1. In the representation given in Proposition 2.3.1, let the distinct primes be q 1, q 2,...,q r, where q 1 < q 2 < < q r. Suppose that q i appears k i times. Then the canonical form of a is: a = q k 1 1 qk 2 2 qkr r. Proposition Let a be an integer greater than 1. Then a is a perfect square if, and only if, the exponents in its canonical form are all even integers. Example Let a = 3, b = 9 and c = Here are the representations specified in Theorem and Definition 2.3.3: a = 3 = 3 1, b = 3 3 = 3 2, c = = Exercise Determine the canonical form of the integer 23!. 2.4 Linear Equations Remark In this section, we assume that a, b and c are integers, with a 0 and b 0. Set d = gcd(a, b). The object is to investigate the solutions, if any, of the equation ax + by = c. Here, x and y are required to be integers. A positive solution is one for which both x and y are positive integers. Proposition The equation in Remark has a solution if, and only if, c is a multiple of gcd(a, b). Proposition Suppose, in Remark 2.4.1, that d = 1 and that one solution is given by: x = x 0, y = y 0. Then the general solution of the equation in Remark is: x = x 0 + bt, y = y 0 at, where t ranges over Z.
4 16 June 25, :53 Chapter 2 Divisibility Proposition Suppose, in Remark 2.4.1, that one solution is given by: x = x 0, y = y 0. Then the general solution of the equation in Remark is: where t ranges over Z. x = x 0 + b d t, y = y 0 a d t, Exercise Find the general solution of the equation 55x + 65y = Find all positive solutions. Remark An equation in which the parameters and unknowns are all integers is called a Diophantine equation, named for the Greek mathematician Diophantus. 2.5 The Euclidean Algorithm Remark Suppose that a and b are positive integers. The purpose here is to see how to determine gcd(a, b) and to find integers x and y such that ax + by = gcd(a, b). Definition Let a and b be integers, with a > b > 0. The Euclidean Algorithm is a procedure for finding gcd(a, b). A list of integers a 1, a 2,..., a n 1, a n is constructed as follows. Let a 1 = a and a 2 = b. Let a 3 be the remainder left by a 1 when divided by a 2, let a 4 be the remainder left by a 2 when divided by a 3, and so on. The process stops when the next term would be 0. Hence, all terms are positive and a n 1 is a multiple of a n. Proposition For the list in Definition it is true that gcd(a 1, a 2 ) = a n. Definition If a, b, c and d are integers, set a b c d = ad bc. This is called a determinant. Definition In these notes, a Euclidean array is an array of integers of the form a 1 a 2 a n 1 a n q 2 q n 1. b 1 b 2 b n 1 b n It is assumed that n 3, and that, for 1 < k < n, a k 1 = a k q k + a k+1 and b k 1 = b k q k + b k+1 Proposition In Definition 2.5.5, either a 1 a 2 b 1 b 2 = a n 1 a n b n 1 b n or according to whether n is even or odd. a 1 a 2 b 1 b 2 = a n 1 b n 1 a n b n,
5 Section 2.6 An Example June 25, : An Example The two integers of interest here are 275 and 635. (1) To find the greatest common divisor of 275 and 635, apply the Euclidean Algorithm: That is, 635 leaves a remainder 85 when divided by 275, and so on. Hence, (2) The object now is to solve the equation gcd(275, 635) = x + 635y = gcd(275, 635). A Euclidean array will be constructed. The second row consists of the quotients found, but not shown, when applying the Euclidean Algorithm: The third row is constructed from right to left. The best way to start is to put 1 and 0 beneath 20 and 5. The first and third rows are related to the second row in exactly the same way, even though they are constructed in opposite directions. The array is: That is, 30 leaves a remainder 4 when divided by 13, and so on. The computations are: 4 = , 13 = , 30 = The first row has an odd number of entries. By Proposition 2.5.6, = Hence, = 5. That is, 275 ( 30) (13) = 5. (3) We now consider the equation By Proposition 2.4.4, the general solution is: where t ranges over Z. 275x + 635y = gcd(275, 635). x = t, y = 13 55t,
6 18 June 25, :53 Chapter 2 Divisibility 2.7 Problems Problem Let q 1 = 2. For n 2, let q n be the smallest prime divisor of the integer 1 + q 1 q 2 q n 1. Show that the q k are distinct, and thereby give another proof of Theorem Verify that the first five terms are 2, 3, 7, 43, 13. Find the next two terms. Problem Determine all ways to make $2.35 with dimes and quarters. Problem Five hundred ducats were used to buy one hundred animals of three types. The burros cost 11 ducats each, the camels cost 9 ducats each, and the dogs cost 2 ducats each. How many animals of each kind were there? Problem Find the general solution of: 2669x y = gcd(2669, 3825). Problem Find the general solution of: 3409x y = 1. Problem Solve: 1234x 4321y = 1. Problem Solve: 4275x y = gcd(4275, 2983). 2.8 Projects Project By Proposition 1.4.8, the product of any two consecutive integers is a multiple of 2. Prove that the product of any three consecutive integers is a multiple of 6. What can you say about the product of any four consecutive integers? Project Suppose that a and b be relatively prime integers, both greater than 1. Look into the question of for which integers c the equation ax + by = c has a positive solution. It will help to deal with three cases c < a + b, c > ab, a + b c ab. Project In addition to the representations for integers given in Theorem and Definition 2.3.3, there is a third method that is useful when dealing with more than one integer at the same time. For n 1, let p n be the n-th prime. For example, p 1 = 2 and p 2 = 3. Any positive integer can be expressed using non-negative powers of the first n primes, for sufficiently large n. For example, to represent the integer c in Example using the first eight primes, we write = Use such representations to illuminate some of the propositions and definitions in Sections 2.1 and 2.2. Assume that all the integers being considered are positive.
7 Section 2.9 Proofs and Suggestions June 25, : Proofs and Suggestions Proof of Theorem (1) Let a and b be integers, not both zero. (2) Let S = {z : z > 0, there exist x and y such that z = ax + by}. (3) The integer a a + b b is an element of S, and so S is nonempty. (4) 0 is a lower bound for S. (5) By Axiom 16, there exists a least integer in S. Call it c. By (2), c is positive. (6) There exist integers x and y such that c = ax + by. (7) By Proposition 1.4.1, there exist integers q and r such that a = qc + r and 0 r c 1. (8) Let x = 1 qx and y = qy. (9) r = a qc = a q(ax + by) = a(1 qx) + b( qy). (10) r = ax + by and 0 r < c. (11) r = 0, since c is the least integer in S. (12) c divides a. (13) Similarly, c divides b. (14) c is a common divisor of a and b and c > 0. (15) Any common divisor of a and b divides ax + by. (16) c = gcd(a, b). Suggestion for Proposition Use Theorem Suggestion for Proposition Let m = lcm(a, b), and let c be a common multiple. Show that if r is the remainder left by c when divided by m, then r is also a common multiple, and must be 0. Suggestion for Proposition Notice that ab d = a b d = a d b. Suggestion for Proposition Let k = ab m. Verify that a = k m b and b = k m a. Suggestion for Proposition Let d = gcd(a, b) and m = lcm(a, b). Use Proposition to show that ab ab m. Use Propositions and to show that d m d.
8 20 June 25, :53 Chapter 2 Divisibility Lemma 2.2.4: Use Theorem to write ax + by = 1, and multiply through by c. Suggestion for Proposition Use Theorem Suggestion for Theorem If p divides ab but not a, apply Lemma and Proposition Proof of Proposition (1) Let p be a prime and let a 1, a 2,..., a n be integers. Suppose that p a 1 a 2 a n. (2) Consider the set S = { j {1, 2,..., n} : p a j a j+1 a n }. By (1) 1 S. Clearly x n for each x S. Thus S is non-empty and bounded above. Therefore S has a maximum. Put m = max S. (3) Now I shall prove that p a m. There are two cases: Case 1 m = maxs = n. Then n S. By the definition of the set S, I conclude that p a n. Case 2 1 m < n. Then m+1 n and m+1 S. Therefore p a m+1 a m+2 a n. Since m S, I conclude that p a m a m+1 a n. Since Euclid s Theorem implies that p a m. p a m a m+1 a n and p a m+1 a m+2 a n, Suggestion for Proposition Use Proposition Suggestion for Proposition Use Proposition Proof of Theorem , Euclid s proof by contradiction. (1) Assume that there exist only finitely many primes, say p 1, p 2,...,p n. (2) Let b = 1 + p 1 p 2 p n. (3) By Proposition 1.6.4, there exists a prime p that divides b. (4) By Proposition , p does not equal any of the p k. (5) That contradicts (1).
9 Section 2.9 Proofs and Suggestions June 25, :53 21 Proof of Proposition (1) Let a Z and a > 1. (2) Put S = { b : b Z, b > 1, b a, b = q 1 q k, where k is a positive integer and q 1,...,q k are primes }. (3) By (1) and Proposition there exists a prime p that divides a. Clearly p S. Thus S. (4) If x S, then x > 1 and x a. Therefore x a. Thus a is an upper bound for S. (5) The lines (3) and (5) imply that Proposition can be applied to the set S and we conclude that S has a maximum. Let c = maxs. (6) Since c S, we conclude that c a. (7) Let x be an integer in S such that x < a. Then a = xs for some integer s. Since x < a, we conclude that s > 1. (8) By Proposition and (7) we conclude that there exist a prime r such that r s, that is s = rt. (9) From (7) and (8) we conclude that a = xrt. Therefore xr a. (10) Since x S there exist primes u 1,..., u j such that x = u 1 u j. Therefore xr = u 1 u j r. (11) Since clearly xr > 1, the lines (9) and (10) imply that xr S. (12) Since r > 1 and x > 1, we have xr > x. (13) The results from the lines (7) through (12) can be summarized as: Let x S. If x < a, then x is not a greatest integer in S. (14) The contrapositive of the statement in (13) is Let x S. If x is a greatest integer in S, then x a. (15) Since c is a greatest integer in S we conclude from (14) that c a. (16) The lines (6) and (15) imply that c = a. (17) Since a = c and c S we conclude that a S. By (2) it follows that there exist primes p 1,...,p n such that a = p 1 p n.
10 22 June 25, :53 Chapter 2 Divisibility Suggestion for Theorem In view of Proposition 2.3.1, the only thing to be proved is the uniqueness of the representation. Let a be an integer greater than 1. Let m, n N and assume that p 1, p 2,...,p m and q 1, q 2,...,q n are primes such that p 1 p 2 p m, q 1 q 2 q n, a = p 1 p 2 p m, a = q 1 q 2 q n. (1) Let k {1, 2,..., n}. Since q 1, q 2,...,q n are primes, the following implication holds: If k < n, then q 1 q 2 q k < a. The contrapositive of this implication is: If q 1 q 2 q k = a, then k = n. (2) We need to prove that m = n and p 1 = q 1,...,p m = q m. Notice that by Proposition we have p 1 = q 1. (3) Assume that m n and consider the set S = { j {1, 2,..., m} : p j q j }. By Proposition / S. We will prove by contradiction that S =. Suppose that S. Since 1 is a lower bound for S, the set S has a minimum. Set k = min S. Since 1 / S, 1 < k m. Then 1,..., k 1 / S, that is p 1 = q 1,..., p k 1 = q k 1. By the assumption a = p 1 p k 1 p k p m = q 1 q k 1 q k q n. Therefore, p k p m = q k q n. Set b = p k p m = q k q n. Since k m, b > 1. By Proposition applied to b we deduce that p k = q k. Therefore k / S. This contradicts k = min S. (4) By (3), S =. Therefore, p 1 = q 1,..., p m = q m. Hence a = p 1 p m = q 1 q m. By (1), a = q 1 q m implies m n. Since we assumed that m n, this implies m = n. Suggestion for Exercise Divide by 5 and try various values of x and y. Proof of Proposition (1) If n = 2, then gcd(a 1, a 2 ) = a 2. (2) Suppose that n > 2. (3) Let q k be such that a k 1 = a k q k + a k+1 for k = 2, 3,..., n 1. (4) The common divisors of a k 1 and a k are just the common divisors of a k and a k+1. (5) gcd(a k 1, a k ) = gcd(a k, a k+1 ) for each k. (6) gcd(a 1, a 2 ) = gcd(a 2, a 3 ) = = gcd(a n 1, a n ) = a n. Suggestion for Proposition Verify that a k 1 b k 1 a k b k = a k a k+1 b k b k+1 for k = 2, 3,..., n 1.
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